11.2. Z VALUE

The standardized normal distribution table (Figure 11-7) is one way to get probability values to use on any normal process or set of data. The Z in the table refers to the number of sigma to the right of the center. The probabilities refer to the area to the right of the Z point.

Be aware that some tables (and computer programs) use different reference points, so examine tables and computer programs carefully before using. Satisfy yourself that you can find data points on the standardized normal distribution table (Figure 11-7) relating to the previous shaft histogram with 0.0001" bins (Figure 11-5).

So there is no confusion reading this chart, let's be sure that it agrees with our reference number of 2/3 (68%) of data points being within ±1 sigma. Looking at the table, with a Z = 1.00 (which means a sigma of 1), we get P = 0.1587, or approximately 0.16. This is illustrated below (Figures 11-8 and 11-9).

Since the left side is a mirror image of the right, this means:

Given that the area under the curve always equals 1 (the sum of all the probabilities equals 1), we know that the white area under the curve = 1 – the shaded tails. This is 1 – (16% + 16%) = 1 – 32% = 68%. This confirms our reference number of 68% (or 2/3, which is easy to remember).

Problem #1

In the shaft process previously discussed, what is the probability of finding a shaft at least 2 sigma (0.0020") over 1.000" in diameter?

(The Z value is an indication of how many sigma, so in this case Z = 2.)

Looking at the standardized normal distribution table (Figure 11-7), Z = 2, P = 0.02275.

Answer: P = 0.02275, or 2.28%.

Problem #2

What is the probability of finding a shaft not greater than 1.002"?

We first must realize that 1.002" is 2 sigma above nominal (since sigma = 0.001"), so Z = 2. Using the standardized normal distribution table (Figure 11-7) to get the probability, looking at Z = 2 we see that P = 0.02275.

Looking at the normal distribution curve at the top of the table (Figure 11-7), we can see that this P is the probability of being greater than 1.002". Since we want not greater than1.002", we must subtract 0.02275 from 1.0000. Again, we know to do this because the total area under the curve, which represents all probabilities, = 1. So, 1 – 0.02275 = 0.97725.

Answer: P = 0.97725, or 97.725%.

Assuming Normal Distribution Use plotted data to visually see if the data is normally distributed. When in doubt, plot more data. Unless the data is dramatically non-symmetrical (data extremely off to one side) or dramatically bimodal (two lobes), assume a normal distribution. The data must clearly show a different distribution or we assume it is normal. There are mathematical formulas to test whether data is normal, but optical inspection of the plotted data is generally sufficient. As you will see later, having a normal distribution allows you to directly use the absolute probability values in the Standardized Distribution Table. However, if the distributions are not normal, the table values can still be used for comparison purposes. Chapter 14 will show that, as long as distributions have similar plots, they can be compared with each other even if their distributions are not normal.

For reference, below is a histogram example that I would consider normal in that we can use the standardized distribution table (Figure 11-7). The distribution, although not perfectly bell-shaped, is not skewed enough to be a concern.

The plot above (Figure 11-10) is based on 48 data points. When in doubt, you can always plot more data.

As you will see in Chapter 14, most of the work we do in Six Sigma does not require a perfectly normal distribution, since we are generally looking for relative change.

Case Study: Incur Great Costs Rather Than Plotting Ahigh-volume consumer product, with over $50,000,000 per year sales, had a historical increase of sales at 3% a year. This growth rate was expected to slowly decrease because of competing products having a longer life. To everyone's surprise, the sales actually increased 13% within a year. Many people had different theories as to why this happened, but no one had any supporting data. The theories for the increased sales ranged from a sudden need by consumers for more of this specific product to an excellent marketing campaign. Plans were started to expand production facilities to support these sales, since the company was having great difficulty meeting this unexpected demand. The marketing of this product was divided into two units, one of which handled large-volume outlets and the other handling mom-and-pop small stores. After over a year of the increased sales and almost $500,000 spent on expansion design, someone noticed that the increased sales had hit the large-volume outlets well before it had affected the low-volume outlets. Since both kinds of outlets served similar customers, this was mysterious. Finally, someone attributed the cause of the increased sales to a design change that had been implemented in the product sometime earlier. This design change inadvertently reduced average product life 10%. Since the large-volume outlets used just-in-time inventories, their customers experienced the effect of the shorter life far earlier than the small outlets, whose inventory usually covered many months of sales. So, the high-volume outlets felt the increased sales level well before the small outlets. The design change was reversed—and the unexpected sales increase disappeared. If someone had just plotted the sales from these two marketing units when the sales increase first appeared, they would have spotted the difference between the two plots, which previously had shown the same 3% growth rate. This would have triggered a more extensive cause analysis one year earlier, prevented a $5,000,000 excessive cost to consumers, and a loss of some customers. Although the supplier had a short-term windfall from the increased sales, it lost much long-term business because it could not supply on a timely basis.

This case study shows how people are quick to react with solutions (expand facilities) at great costs, but will spend little time on plotting data and truly understanding a root cause.

Normal Distribution Symmetry Remembering that each half of the normal distribution curve is a mirror image of the other, we can use data given for the plus side to solve problems related to both sides.

Problem #3

In the shaft process previously discussed, what is the probability of getting a shaft below 0.9978" in diameter?

This shaft diameter is 0.0022" below nominal (1 – 0.9978"). Since sigma = 0.0010", this is 2.2 sigma below nominal, so Z = 2.2. Looking at the standardized normal distribution table (Figure 11-7), Z = 2.2, we see that P = 0.0139. 1.39% of the data points would occur above a positive 2.2 sigma. Since the negative side of the probability table is a mirror image of the positive side, the probability also applies to a negative 2.2 sigma.

Answer: P = .0139, or 1.39%.

No Values Occur Exactly at a Z Point In using a standardized normal distribution curve, all values are assumed to be above or below a Z point. For example, if you wanted to know what percent of values are "above Z = 2," it would be the same as the percent of values "at or above Z = 2."

For simplicity, the previous shaft data had a sigma = 0.0010". This was to make calculations and understanding easier. Usually the sigma doesn't correlate with the bin edges, nor is it such an even number. This in no way changes the logic or diminishes the value of the standardized normal distribution table (Figure 11-7), as illustrated below.

Problem #4

Using the shaft example, let's assume that the customer has complained that the amount of variation in the shafts is causing him process problems. The customer is especially critical of shafts less than 0.9980" and greater than 1.0020" (more than 0.0020" from nominal). In response, the lathe is overhauled. On taking another 1000 measurements, it is determined that the average has stayed at 1.0000", but the sigma has been reduced from 0.0010" to 0.0007".

The reduced sigma means that the variation among shafts is less than before the overhaul. We want to communicate to the customer what improvement he can expect in future shipments, specifically what reduction he will see in shafts more than 0.0020" above or below the nominal 1.0000" diameter.

Before the overhaul (Problem #1, sigma = 0.0010"), we found that the probability of finding a shaft at least 0.0020" above 1.0000" in diameter was 0.02275. Given that both sides of the curve are mirror images, we double that number to calculate the chances of being at least 0.0020" ± nominal.

P = 0.02275 * 2 = 0.0455, or 4.55% (P before the overhaul).

We must now calculate the P with the new reduced sigma (0.0007"). First, we see how many sigma "fit" between nominal and 0.0020". We use the plus side first, since that is the data given to us in the standardized normal distribution table (Figure 11-7).

0.0020" / 0.0007" = 2.86 sigma fit! This gives us the Z to use in the standardized normal distribution table (Figure 11-7).

Using the standardized normal distribution table (Figure 11-7), looking at Z = 2.85 (the closest table data point), the P value we read from the table is 0.002186. So, 0.2186% of the shafts will be at least 0.0020" above nominal. We double this to include those at least 0.0020" below 1.0000" diameter.

2 * 0.002186 = 0.004372

The total P is 0.004372, or 0.4372% (P after the overhaul).

Answer: Since the process had been making 4.55% at 0.0020" above or below 1.0000" and it is now making 0.4372%, the customer can expect to see 9.6% (0.437 / 4.55) of the former problem shafts.

Problem #5

Let's change the above problem again to make it even more "real." After the overhaul, the lathe sigma is reduced to 0.0007" (same as above), but the average shaft diameter is now 1.0005". The process plot is still normal. Will the customer be receiving fewer problem shafts than before the overhaul?

Since the process average is no longer centered at the 1.0000" nominal, the amount of product outside the 0.9980"-to-1.0020" target is different for the large diameters than for the small diameters, so calculate each independently.

First, we will calculate the P for the too-large shafts. As before, we see how many sigma (0.0007") "fit" between the new process average (1.0005") and the +1.0020" upper limit.

This calculation is (1.0020" – 1.0005") / 0.0007" = 2.143 sigma fit.

Looking at the standardized normal distribution table (Figure 11-7), we see that the P at a Z of 2.15 (closest value to 2.143) is 0.01578. That means that 1.578% of the shafts will be 1.0020" in diameter or larger.

Looking at the too-small shafts, we do a similar calculation. First, find the value for the difference between the process average and the lower end of the target. (The process average is 1.0005" and the lower target value is 0.9980".) The difference is 1.0005" – 0.9980" = 0.0025".

We then see how many sigma (0.0007") "fit": 0.0025" / 0.0007" = 3.57 sigma. Although we must use the data on the positive end of the curve, we know that the mirror image would be identical. Looking at the P value for a Z of 3.55, we get P = .0001927. So, 0.019% of the shafts will be .9980" or smaller.

Answer: When we add the too-large and too-small diameter shafts, we get 1.578% + 0.019% = 1.60% of the shafts will be at least 0.0020" off the 1.0000" nominal. 1.60% is less than the 4.45% the customer was receiving before the overhaul, so the customer will be receiving a better product. Note, however, that 1.60% is much higher than the 0.4372% (Problem #4) the customer would receive if the process were centered.

This change in both the average and the sigma is not unusual in a process change. However, it is usually not difficult to get the process mean back to the target center (in this case, 1.0000" diameter). If the process center is put back to nominal, we get the 10-fold improvement we saw in the earlier problem.

Adjusting a Process's Mean Versus Reducing Its Sigma Normally moving a process's mean is easier than trying to reduce its sigma. A mean change is often just choosing the center around which the process will be run; it requires no major process change. A sigma reduction often requires a significant change in the process itself, like dramatically slowing the process or changing the equipment being used.

Note that in the above cases, the standardized normal distribution table (Figure 11-7) Z values that were used were those closest to the calculated values of Z. There was no attempt to extrapolate or go to another table or computer program for greater accuracy. Either would have been possible, but if you look at the relative values obtained versus the changes being noted, the greater accuracy was not required. Often the calculation accuracy far exceeds the requirements of the output results.

Using Excel to Get Normal Distribution Values Those wishing to use the computer to get the probability values for various values of Z can use Excel. After bringing up the Excel worksheet, click on in the toolbar. Under "category," click on "statistical." Then, under "function," click on "NORMSDIST." When you enter a Z value, it gives you the probability values using the left end of the distribution as the reference zero, whereas the standardized normal distribution table (Figure 11-7) uses the right end of the distribution as zero. To convert either one to the other, simply subtract the value from 1. For example, if you enter Z = 2 in the Excel NORMSDIST, you get a probability value of 0.97725. This is the probability of being less than the Z value. 1.00000 – 0.97725 = 0.02275 is the probability of being greater than Z, which matches the probability given for Z = 2 on the standardized normal distribution table (Figure 11-7).

Just for information purposes, the 6-sigma process is sometimes referred to as three defects per million. If you look at the standardized normal distribution table (Figure 11-7) you will see that three defects per million is 4.65 sigma, not 6 sigma. The 6-sigma short-term target is tighter than 4.65 sigma, because it assumed a process drift would take place over time. If you started with a process that was 6 sigma short term, the goal was to have a 4.65-sigma process when the long-term drift was included.